Solve the given differential equation subject to the indicated initial condition. `(e^(-y) + 1)sinx dx = (1 +cos x ) dy , y(0) = 0`

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Solve the given differential equation subject to the indicated initial condition.  

`(e^(-y) + 1)sinx dx = (1 +cos x ) dy , y(0) = 0`

Solution:

To solve the differential equation `(e^(-y) + 1)sinx dx = (1 +cos x ) dy` with the initial condition `y(0) = 0`, we can first use separation of variables to get:

`∫ (e^(-y) + 1) dy = ∫ (1 +cos x ) dx`

Integrating both sides with respect to their respective variables gives:

`e^(-y) = x + sin x + C`

Solving for y gives:

`y = -ln(x + sin x + C)`

Now we can use the initial condition to find the value of C:

`y(0) = -ln(0 + sin 0 + C) = 0`

`C = -1`

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So the general solution is:

`y = -ln(x + sin x - 1)`

And the solution to the differential equation subject to the initial condition is:

`y = -ln(x + sin x - 1)`